Question

$a^2+\frac{1}{a^2}=2$ then $\frac{1}{a}+a$=?

Answer 1

Step-1: Find ${\left(\frac{1}{a}+a\right)}^2$

${\left(\frac{1}{a}+a\right)}^2$$=a^2+\frac{1}{a^2}+2$ ( As ${\left(a+b\right)}^2=a^2+b^2+2ab$ )

${\left(\frac{1}{a}+a\right)}^2$$=2+2$ ( As $a^2+\frac{1}{a^2}=2$is given)

${\left(\frac{1}{a}+a\right)}^2$$=4$

Step-2: Find $\frac{1}{a}+a$

To find $\frac{1}{a}+a$take square root on both sides of the equation- (i) ;${\left(\frac{1}{a}+a\right)}^2$$=4$

$\sqrt{{\left(\frac{1}{a}+a\right)}^2}$$=\sqrt{4}$

$\frac{1}{a}+a$$=2$

Hence, $\frac{1}{a}+a$is $2$